voting game combinatorics Let $N=\{1, \ldots, n \}$ be a set of voters and $A=\{c_1, \ldots, c_m \}$ be a set of candidates of an election. For each voter $i$, we assign a strict linear order $\succ_i$ such that $c_a \succ_i c_b$ if $i$ prefers $c_a$ over $c_b$. 
We compute a Borda's score $S(c_a)$ to a candidate $c_a$, as follows:
$c_a$ receives $m-1$ points for each voter who ranks it first, $m-2$ points for each voter who ranks it second, and so forth. The Borda's score is the sum of all these points.
I want to show that 
$$S(c_a) = \sum_{c_b \in A \setminus \{c_a \}} | \{
 i \in N : c_a \succ_i c_b \} |.$$
Here is what I get:
\begin{eqnarray}
RHS & = & \sum_{c_b \in A \setminus \{c_a \}} \sum_{p=0}^{m-1} | \{
 i \in N : c_a \succ_i c_b \text{ and there are } p \text{ elements less preferred by } i \text{ to } c_b\} | \\
& = & \sum_{p=0}^{m-1} \sum_{c_b \in A \setminus \{c_a \}} \{
 i \in N : c_a \succ_i c_b \text{ and there are } p \text{ elements less preferred by } i \text{ to } c_b\} |\\
& = & \sum_{p=0}^{m-1} (\text{Total number of voters ranking } c_a \text{ as } (m-p)\text{th } ),
\end{eqnarray}
which doesn't seem to be true. Any mistakes?
 A: I’ll write $S_i(c)$ for voter $i$’s contribution to $S(c)$. With that notation the first couple of steps of your calculation can be written
$$\begin{align*}
\sum_{c\in A\setminus\{c_a\}}|\{i\in N:c_a\succ_i c\}|&=\sum_{c\in A\setminus\{c_a\}}\sum_{p=0}^{m-1}|\{i\in N:c_a\succ_i c\text{ and }S_i(c)=p\}|\\\\
&=\sum_{p=0}^{m-1}\sum_{c\in A\setminus\{c_a\}}|\{i\in N:c_a\succ_i c\text{ and }S_i(c)=p\}|\;,
\end{align*}$$
and you’re claiming that
$$\sum_{c\in A\setminus\{c_a\}}|\{i\in N:c_a\succ_i c\text{ and }S_i(c)=p\}|\tag{1}$$
is the number of voters ranking $c_a$ as $(m-p)$-th, i.e., the number of voters $i$ such that $S_i(c_a)=p$. That, however, isn’t true. Suppose that there are $3$ candidates, and every voter ranks $c_a$ first and $c$ last. Then for $p=0$ expression $(1)$ evaluates to $3$, but in fact no voter ranks $c_a$ last, or even second.
Instead of introducing a summation on $p$, introduce one on voters:
$$\begin{align*}
\sum_{c\in A\setminus\{c_a\}}|\{i\in N:c_a\succ_i c\}|&=\sum_{c\in A\setminus\{c_a\}}\sum_{i\in N}[c_a\succ_i c]\\\\
&=\sum_{i\in N}\sum_{c\in A\setminus\{c_a\}}[c_a\succ_i c]\\\\
&=\sum_{i\in N}S_i(c_a)\\\\
&=S(c_a)\;,
\end{align*}$$
where the square brackets are Iverson brackets.
